# Physics problem

CM123
Posers and Puzzles 03 Aug '09 07:43
1. 03 Aug '09 07:43
*Basic Physics knowledge required.

Prove that a semicircle (radius R) with a plank (length L, mass M) rocking on the semicircle has 'simple harmonic motion' (F=-KX, k is constant, X is distance from equilibrium position). Ignore friction, gravity=9.8m/s^2.

This is all the information I was given, been trying it for ages๐
Baby Gauss
03 Aug '09 12:35
Originally posted by CM123
*Basic Physics knowledge required.

Prove that a semicircle (radius R) with a plank (length L, mass M) rocking on the semicircle has 'simple harmonic motion' (F=-KX, k is constant, X is distance from equilibrium position). Ignore friction, gravity=9.8m/s^2.

This is all the information I was given, been trying it for ages๐
I'm not sure if I understtod the text of the problem... Can you post a picture or maybe worth it a little differently..?
3. 04 Aug '09 05:32
Can't figure how to post pictures....

A plank is rocking on top of circular object, at any time in its motion, the force acting towards equilibrium position can be described as being equal to a constant times the extension from the equilibrium position. Show what this constant is.

The equilibrium position is when the plank is at rest on the object, and the extension is the distance from the equilibrium position to any point along the motion.

The difficult part is that the pivoting point on the plant changes as it rotates.

If you want a diagram send me a message with your email.
4. 04 Aug '09 12:38
OK, first point - you can't ignore friction! The plank will just slide off otherwise.

This was my approach. Could be horribly wrong.

First, definitions. Let a be the angle between the vertical and the point of contact. X is the distance between the centre of the plank and the point of contact. So X = Ra.

My approach is to think about it as rotation about the point of contact.
- Gravity = Mg
- Component of gravity in the direction perpendicular to the plank = Mg cos a
- Moment = Mg cos a x X = MgRa cos a

The moment of inertia of a plank about the centre is ML^2/12
The moment of inertia of a plank about a point X from the centre is ML^2/12 + MX^2

Therefore, using Moment = Moment of Inertia x Angular acceleration:

M(L^2/12 + R^2a^2) x d2a/dt2 = - MgRa cos a

d2a/dt2 = - gRa cos a/(L^2/12 + R^2a^2)

Assuming a small oscillation (it isn't SHM otherwise), linearise this with respect to a:

d2a/dt2 = -(12gR/L^2)a

[I think there are probably some approximations involved in this, but they're all on the same scale as the linearisation anyway. I've just tried hitting it with a Lagrangian mechanics sledgehammer, and I get the same answer to the first order.]
5. 05 Aug '09 05:40

Looks good at a brief glance, I haven't had much time to go over it - two questions however...

How did you calculate the moment of inertia, and can gravity have a component perpendicular to the plank when it is not at rest?

Also, I think SHM can have large oscillations, just not in this case.
6. 05 Aug '09 10:00
Originally posted by CM123

Looks good at a brief glance, I haven't had much time to go over it - two questions however...

How did you calculate the moment of inertia, and can gravity have a component perpendicular to the plank when it is not at rest?

Also, I think SHM can have large oscillations, just not in this case.
I looked up the moment of inertia of a rod (ML^2/12) about the centre, then used the parallel axis theorem to calculate it about the pivot point (I' = I + Mx^2).

I don't think there are any problems with using the component of gravity. There is an approximation because I've neglected the fact that the pivot point is moving, but that comes out as O(a^3) I think.

Yes, SHM can have large oscillations - in that case the equation of motion would have been exact, and I wouldn't have had to approximate.
7. 05 Aug '09 10:08
Since I worked it out...here's the sledgehammer. (This took me back - I've not used this for years!). I can't guarantee that there are no errors, but it does end up with the same answer.

Position of point of contact (relative to the centre of the semi-circle) is [R sin a, R cos a]

Position of Centre of Mass C = [R sin a - X cos a, R cos a + X sin a]

=> C = R[sin a - a cos a, cos a + a sin a]
=> dC/dt = Ra[sin a, cos a]da/dt

I'm going to write a' = da/dt for convenience.

Kinetic Energy = Rotational KE + Translational KE
= Ia'^2/2 + MV^2/2
= (ML^2/12)a'^2/2 + MR^2a^2a'^2/2
= Ma'^2/2(L^2/12 + R^2a^2)

Potential energy (relative to centre of semi-circle) = MgR(cos a + a sin a)

Langrangian = KE - PE
= Ma'^2/2(L^2/12 + R^2a^2) - MgR(cos a + a sin a)

Hit this with the Euler-Lagrange equation, and our equation of motion is:

d/dt[Ma'(L^2/12 + R^2a^2)] - MR^2a'^2a + MgRa cos a = 0

=> a''(L^2/12 + R^2a^2) + 2R^2a'^2a - R^2a'^2a + gRa cos a = 0

=> a'' = -aR[R a'^2 + g cos a]/[L^2/12 + R^2a^2]

This is the _exact_ equation of motion. Which shows that my previous approach is missing an O(a^3) term. But linearising gives the same result.
8. 07 Aug '09 13:40
Thanks for that ๐